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Math Help - Inequlity

  1. #1
    Junior Member
    Joined
    Dec 2009
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    33

    Inequlity

    Let a>0 and k>1. Prove that for comlex z

    |arg\ (z)|\leq 2\,\arccos\frac 1k \Rightarrow |a+z|\geq \frac {a+|z|}k

    Hint: \cos x is decreasind in [o,\pi]. Not so difficult, I think?
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  2. #2
    Member
    Joined
    Mar 2009
    Posts
    91

    Inequality

    How about this?

    Let \theta=\arg z. Since |\arg z|\leq 2\arccos\dfrac1k=2\alpha and \cos x is decreasing on [0,\pi] we have

    \cos\theta\geq\cos 2\alpha=2\cos^2\alpha-1=\dfrac2{k^2}-1.

    Thus

    \mathop{\textrm{Re}} z=|z|\cos\theta\geq\left(\dfrac2{k^2}-1\right)|z|.

    Now
    <br />
|a+z|^2=a^2+2a\mathop{\textrm{Re}}z+|z|^2\geq a^2+2a|z|\left(\dfrac2{k^2}-1\right)+|z|^2

    Hence
    <br />
|a+z|^2-\dfrac{(a+|z|)^2}{k^2}\geq\left(1-\dfrac1{k^2}\right)(a^2-2a|z|+|z|^2)

    Since a^2-2a|z|+|z|^2=(a-|z|)^2\geq 0 and k>1 the result follows.
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